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Question Number 110939 by ajfour last updated on 31/Aug/20

Commented by ajfour last updated on 31/Aug/20

$I f b o t h e l l i p s e s a r e c o n g r u e n t, f i n d$ $\tan α .$
	Answered by ajfour last updated on 01/Sep/20

	Answered by mr W last updated on 01/Sep/20

	Commented by mr W last updated on 02/Sep/20

	Commented by ajfour last updated on 02/Sep/20
	$let P(p,−q) (p^2 /a^2 )+(q^2 /b^2 )=1 .......(3) tangent at P ((px)/a^2 )+((−qy)/b^2 )=1 slope=((b^2 /a^2 ))((p/q)) = tan γ (({(p+h)^2 +(−q+k)^2 }cos^2 α)/a^2 ) +(({(p+h)^2 +(−q+k)^2 }sin^2 α)/b^2 )=1 ⇒ (p+h)^2 +(−q+k)^2 =((a^2 b^2 )/(b^2 cos^2 α+a^2 sin^2 α)) ..........(4) And (u^2 /a^2 )+(v^2 /b^2 )=1 (eq. of lower ellipse in u-v axes) say CP^( 2) =s^2 =(p+h)^2 +(−q+k)^2 eq. of tangent at P ((uscos α)/a^2 )+((vssin α)/b^2 )=1 slope = −((b^2 /a^2 ))(cot α) = −tan β β+γ=φ ⇒ tan^(−1) (((b^2 cos α)/(a^2 sin α)))+tan^(−1) (((b^2 p)/(a^2 q)))=φ ⇒ (p/q)=(a^2 /b^2 ){((λ−(b^2 /(a^2 tan α)))/(1+((λb^2 )/(a^2 tan α))))} .....(5) scos (α+tan^(−1) λ)=p+h .....(6) ssin (α+tan^(−1) λ)=−q+k .....(7) unknowns are (b/a), λ, h, k, p, q, α . .........$
	$l e t P (p, - q)$ $\frac{p^{2}}{a^{2}} + \frac{q^{2}}{b^{2}} = 1 . . . . . . . (3)$ $t a n g e n t a t P$ $\frac{p x}{a^{2}} + \frac{- q y}{b^{2}} = 1$ $s l o p e = (\frac{b^{2}}{a^{2}}) (\frac{p}{q}) = \tan γ$ $\frac{{{(p + h)}^{2} + {(- q + k)}^{2}} \cos^{2} α}{a^{2}}$ $+ \frac{{{(p + h)}^{2} + {(- q + k)}^{2}} \sin^{2} α}{b^{2}} = 1$ $\Rightarrow$ ${(p + h)}^{2} + {(- q + k)}^{2} = \frac{a^{2} b^{2}}{b^{2} \cos^{2} α + a^{2} \sin^{2} α}$ $. . . . . . . . . . (4)$ $A n d$ $\frac{u^{2}}{a^{2}} + \frac{v^{2}}{b^{2}} = 1 (e q . o f l o w e r e l l i p s e i n$ $u - v a x e s)$ $s a y {C P}^{2} = s^{2} = {(p + h)}^{2} + {(- q + k)}^{2}$ $e q . o f t a n g e n t a t P$ $\frac{u s \cos α}{a^{2}} + \frac{v s \sin α}{b^{2}} = 1$ $s l o p e = - (\frac{b^{2}}{a^{2}}) (\cot α) = - \tan β$ $β + γ = ϕ$ $\Rightarrow \tan^{- 1} (\frac{b^{2} \cos α}{a^{2} \sin α}) + \tan^{- 1} (\frac{b^{2} p}{a^{2} q}) = ϕ$ $\Rightarrow \frac{p}{q} = \frac{a^{2}}{b^{2}} {\frac{λ - \frac{b^{2}}{a^{2} \tan α}}{1 + \frac{λ b^{2}}{a^{2} \tan α}}} . . . . . (5)$ $s \cos (α + \tan^{- 1} λ) = p + h . . . . . (6)$ $s \sin (α + \tan^{- 1} λ) = - q + k . . . . . (7)$ $u n k n o w n s a r e$ $\frac{b}{a}, λ, h, k, p, q, α .$ $. . . . . . . . .$
	Commented by mr W last updated on 02/Sep/20

	$e q n . o f u p p e r e l l i p s e :$ $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1 . . . (i)$ $μ = \frac{b}{a}$ $λ = \tan ϕ$ $C (- h, - k)$ $e q n . o f C A :$ $y + k = (x + h) \tan ϕ$ $\Rightarrow λ x - y + λ h - k = 0$ $e q n . o f C B :$ $y + k = - \frac{x + h}{\tan ϕ}$ $\Rightarrow x + λ y + h + λ k = 0$ $a^{2} λ^{2} + b^{2} = {(λ h - k)}^{2}$ $a^{2} + b^{2} λ^{2} = {(h + λ k)}^{2}$ $\Rightarrow λ h - k = + \sqrt{a^{2} λ^{2} + b^{2}}$ $\Rightarrow h + λ k = \sqrt{a^{2} + b^{2} λ^{2}}$ $\Rightarrow h = \frac{- λ \sqrt{a^{2} λ^{2} + b^{2}} + \sqrt{a^{2} + b^{2} λ^{2}}}{1 + λ^{2}}$ $\Rightarrow η = \frac{h}{a} = \frac{- λ \sqrt{λ^{2} + μ^{2}} + \sqrt{1 + μ^{2} λ^{2}}}{1 + λ^{2}}$ $\Rightarrow k = \frac{λ \sqrt{a^{2} + b^{2} λ^{2}} + \sqrt{a^{2} λ^{2} + b^{2}}}{1 + λ^{2}}$ $\Rightarrow κ = \frac{k}{a} = \frac{λ \sqrt{1 + μ^{2} λ^{2}} + \sqrt{λ^{2} + μ^{2}}}{1 + λ^{2}}$ $e q n . o f l o w e r e l l i p s e :$ $\frac{{[(x + h) \cos ϕ + (y + k) \sin ϕ]}^{2}}{a^{2}} + \frac{{[- (x + h) \sin ϕ + (y + k) \cos ϕ]}^{2}}{b^{2}} = 1$ $\frac{{[(x + h) + (y + k) λ]}^{2}}{a^{2}} + \frac{{[- (x + h) λ + (y + k)]}^{2}}{b^{2}} = 1 + λ^{2} . . . (i i)$ $(i) a n d (i i) s h o u l d t o u c h e a c h o t h e r .$ $l e t x = a \cos θ, y = - b \sin θ f o r p o i n t P$ $\Rightarrow {[η + \cos θ + (κ - μ \sin θ) λ]}^{2} + \frac{{[κ - μ \sin θ - (η + \cos θ) λ]}^{2}}{μ^{2}} = 1 + λ^{2} . . . (i i i)$ $f o r g i v e n μ w e c a n d e t e r m i n e λ$ $s u c h t h a t (i i i) h a s o n l y o n e s o l u t i o n$ $f o r θ .$ $o r f o r g i v e n λ w e c a n d e t e r m i n e μ .$
	Commented by ajfour last updated on 02/Sep/20

	${I s n}^{'} t i t o k a y S i r, 7 e q n s . & 7 u n k n o w n s . .$
	Commented by mr W last updated on 02/Sep/20

	${i t}^{'} s o k a y s i r! l e t m e s e e i f w e c a n$ $s i m p l i f y m o r e .$
	Commented by mr W last updated on 02/Sep/20

	Commented by mr W last updated on 02/Sep/20

	Commented by mr W last updated on 02/Sep/20

	Commented by ajfour last updated on 02/Sep/20

	$t h a n k y o u S i r, b u t f o r μ = 1$ $t h e r e i s n o λ i t h i n k . . .$
	Commented by mr W last updated on 02/Sep/20

	$c o r r e c t!$ $μ_{m a x} \approx 0.38$
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